SO2 reacts with H2S as follows: 2H2S + SO2 –> 3S + 2H2O When 7.50 g of H2S reacts with 12.75 g of SO2, which statement is true?

SO2 reacts with H2S as follows: 2H2S + SO2 –> 3S + 2H2O
When 7.50 g of H2S reacts with 12.75 g of SO2, which statement is true?
A.H2O is the limiting reagent
B.H2S is the limiting reagent
C.SO2 is the limiting reagent
D.S is the limiting reagent

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Answer

• H2S is the limiting reagent
  Explanation:
  2H2S + SO2 –> 3S + 2H2O
  Mole ratio of 2H2S : SO2 = 2 : 1
  Amount of each reagent in moles
  Moles = mass / molar mass
  2H2S
  Moles = 7.5 g / 34 g/mol = 0.22 moles
  Moles of SO2
  12.75 g / 64 g/mol = 0.2 moles
  According to the equation, 2 moles of H2S requires 1 mole of SO2
  Thus, 0.22 moles will require = 0.22 moles * 1 / 2 = 0.11 moles of SO2
  Available moles of SO2 is 0.2, while the amount required by H2S for the reaction is 0.11. This implies that H2S is in excess of (0.2-0.11) 0.09 moles
  The limiting reagent is H2S because it requires less SO2 than what is available